Prove that two angles are equal

Question

$M$ is the midpoint of $BC$ in the triangle $\Delta ABC$. $D$ lies on $AC$, and $AD = BD$. $E$ lies on the line $AM$, $DE$ is parallel to $AB$. How can I prove that the angles $D\hat{B}E$ and $A\hat{C}B$ are equal?

Answer 1

Let $L$ be chosen on the line $AM$ so that $AM = ML$, i.e. $M$ is the midpoint of segment $AL$. Then $ABLC$ is a parallelogram. Thus $AB$ is parallel to $DE$ and $CL$. Let $N$ be the intersection of line $CL$ with line $BD$.

Lemma 1. Since $AD = BD$, the trapezoid $ABCN$ is isosceles with $AN = BC$ and $\angle \, ANB = \angle \, ACB$

Proof: Indeed, the fact that triangle $ABD$ is isosceles implies that $\angle \, ABD = \angle \, BAD = \alpha$. Furthermore, since $AB$ is parallel to $CN$ $$\angle \, NCD = \angle \, NCA = \angle \, BAC = \angle\, BAD = \alpha$$ and $$\angle \, CND = \angle \, CNB = \angle \, ABN = \angle\, ABD = \alpha$$ Consequently, triangle $CDN$ is isosceles with $DC = DN$. Therefore triangles $BDC$ and $ADN$ are congruent so $AN = BC$. Hence $\angle \, ANB = \angle \, ACB$.

Lemma 2. The following equalities gold: $$\frac{DF}{NL} = \frac{BF}{BL} = \frac{AD}{AC} = \frac{AQ}{AN} =\frac{QE}{NL}$$ yielding $QE = DF = AB = CL$.

Proof: Look at triangle $BNL$ and the line $DF$ parallel to $NL$. Then $$\frac{DF}{NL} = \frac{BF}{BL}$$ by the intercept theorem for the pair of lines $BN$ and $BL$ intersected by the two parallel lines $BF$ and $NL$. Alternatively this result follows from the similarity between the triangles $BDF$ and $BNL$. Since quad $ABLC$ is a parallelogram and $DF$ is parallel to both $AB$ and $CL$, the quad $ABFD$ is also a parallelogram so $BF = AD$ and $BL = AC$ which implies
$$\frac{BF}{BL} = \frac{AD}{AC}$$ Next, look at triangle $ANC$ and the line $QD$ parallel to $NC$. Then $$\frac{AD}{AC} = \frac{AQ}{AN}$$ by the intercept theorem for the pair of lines $AN$ and $AC$ intersected by the two parallel lines $QD$ and $NC$. Alternatively this result follows from the similarity between the triangles $AQD$ and $ANC$. Finally, look at triangle $ANL$ and the line $QE$ parallel to $NL$. Then $$\frac{AQ}{AN} = \frac{QE}{NL}$$ by the intercept theorem for the pair of lines $AN$ and $AL$ intersected by the two parallel lines $QE$ and $NL$. Alternatively this result follows from the similarity between the triangles $AQE$ and $ANL$.

Completing the proof of the main result: Since $QE = AB$ and $QE$ is parallel to $AB$, the quad $ABEQ$ is a parallelogram, which means that $BE$ is parallel to $AQ$ and therefore $BE$ is parallel to $AN$ too. This means that $\angle \, DBE = \angle \, NBE = \angle \, ANB = \angle \, ACB$.